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Ta có:
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+16\right)+8\left(x^2+10x+16\right)+16\)
\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)
\(=\left(x^2+10x+16+4\right)^2=\left(x^2+10+20\right)^2\)
k nha!!
\(\text{( x + 2 ) ( x + 4 ) ( x + 6 ) ( x + 8 ) + 16}\)
\(\text{Phân tích thành nhân tử :}\)
\(\left(x^2+10x+20\right)^2\)
\(x^{16}+x^8-2=x^{16}-x^8+2x^8-2=x^8\left(x^8-1\right)+2\left(x^8-1\right)\)
\(=\left(x^8+2\right)\left(x^8-1\right)=\left(x^8+2\right)\left(x^4+1\right)\left(x^4-1\right)\)
\(=\left(x^8+2\right)\left(x^4+1\right)\left(x^2+1\right)\left(x^2-1\right)\)
\(=\left(x^8+2\right)\left(x^4+1\right)\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\)
(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
đặt t=x2+10x+16 ta được:
t.(t+8)+16
=t2+8t+16
=(t+4)2
thay t=x2+10x+16 ta được:
(x2+10x+16)2
=[(x+2)(x+8)]2
=(x+2)2(x+8)2
vậy (x+2)(x+4)(x+6)(x+8)+16 =(x+2)2(x+8)2
(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
đặt t=x2+10x+16 ta được:
t.(t+8)+16
=t2+8t+16
=(t+4)2
thay t=x2+10x+16 ta được:
(x2+10x+16)2
=[(x+2)(x+8)]2
=(x+2)2(x+8)2
vậy (x+2)(x+4)(x+6)(x+8)+16 =(x+2)2(x+8)2
\(x^{16}+x^8-2\)
\(=\left(x^8-1\right)\left(x^8+2\right)\)
\(=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+2\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+2\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+2\right)\)
= (x^16 -1)+(x^8 -1)
= (x^8 -1)(x^8 +1)+ (x^8 -1)
= (x^8 -1)(x^8 +2)
= (x^4 -1)(x^4 +1)(x^8 +2)
= (x^2 -1)(x^2 +1)(x^4 +1)(x^8 +2)
= (x-1)(x+1)(x^2 +1)(x^4 +1)(x^8 +2)
Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
Đề Phân tích đa thức thành nhân tử 1/(1 - x )+ 1/(1+x)+2/(1+x^2)+ 4/(1+x^4)+8/(1+x^8) - 16/(1+ x^16)
x 16 + x 8 − 2 = ( x 8 ) 2 + x 8 − 2 = ( x 8 − 1 ) ( x 8 + 2 ) = ( x 4 − 1 ) ( x 4 + 1 ) ( x 8 + 2 ) = ( x 2 − 1 ) ( x 2 + 1 ) ( x 4 + 1 ) ( x 8 + 2 ) = ( x − 1 ) ( x + 1 ) ( x 2 + 1 ) ( x 4 + 1 ) ( x 8 + 2 )